If $Q_l in ℝ^{mtimes l}$, why does $Q_lQ_l^T in ℝ^{mtimes m}$ have nullity $m-l$?












1












$begingroup$


If $Q_l in ℝ^{mtimes l}$ is a matrix that is the first $l$ columns of $Q in ℝ^{mtimes m}$ where Q is orthogonal, why does $Q_lQ_l^T in ℝ^{mtimes m}$ have nullity $m-l$?



The nullspace of $Q_lQ_l^T$ is {$x: Q_lQ_l^Tx=0$}



Since $Q_lQ_l^T=I$ why is the nullity not $0$?



Also if $A in ℝ^{mtimes n} , mge n$ why is $|Q_l^TA|_2 = |A|_2$ only if $l=m$?



I know that since $Q_lQ_l^T=I implies |Q_l|_2=1$ and so $|Q_l^TA|_2 le |A|_2$






linear-algebra







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    The dimension of the nullspace is also called nullity
    $endgroup$
    – Shubham Johri
    Dec 21 '18 at 18:17










  • $begingroup$
    The nullity of $I$ is $0$
    $endgroup$
    – Shubham Johri
    Dec 21 '18 at 18:23










  • $begingroup$
    Was a typo my bad
    $endgroup$
    – pablo_mathscobar
    Dec 21 '18 at 18:24
















1












$begingroup$


If $Q_l in ℝ^{mtimes l}$ is a matrix that is the first $l$ columns of $Q in ℝ^{mtimes m}$ where Q is orthogonal, why does $Q_lQ_l^T in ℝ^{mtimes m}$ have nullity $m-l$?



The nullspace of $Q_lQ_l^T$ is {$x: Q_lQ_l^Tx=0$}



Since $Q_lQ_l^T=I$ why is the nullity not $0$?



Also if $A in ℝ^{mtimes n} , mge n$ why is $|Q_l^TA|_2 = |A|_2$ only if $l=m$?



I know that since $Q_lQ_l^T=I implies |Q_l|_2=1$ and so $|Q_l^TA|_2 le |A|_2$






linear-algebra







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    The dimension of the nullspace is also called nullity
    $endgroup$
    – Shubham Johri
    Dec 21 '18 at 18:17










  • $begingroup$
    The nullity of $I$ is $0$
    $endgroup$
    – Shubham Johri
    Dec 21 '18 at 18:23










  • $begingroup$
    Was a typo my bad
    $endgroup$
    – pablo_mathscobar
    Dec 21 '18 at 18:24














1












1








1


1



$begingroup$


If $Q_l in ℝ^{mtimes l}$ is a matrix that is the first $l$ columns of $Q in ℝ^{mtimes m}$ where Q is orthogonal, why does $Q_lQ_l^T in ℝ^{mtimes m}$ have nullity $m-l$?



The nullspace of $Q_lQ_l^T$ is {$x: Q_lQ_l^Tx=0$}



Since $Q_lQ_l^T=I$ why is the nullity not $0$?



Also if $A in ℝ^{mtimes n} , mge n$ why is $|Q_l^TA|_2 = |A|_2$ only if $l=m$?



I know that since $Q_lQ_l^T=I implies |Q_l|_2=1$ and so $|Q_l^TA|_2 le |A|_2$






linear-algebra







share|cite|improve this question











$endgroup$




If $Q_l in ℝ^{mtimes l}$ is a matrix that is the first $l$ columns of $Q in ℝ^{mtimes m}$ where Q is orthogonal, why does $Q_lQ_l^T in ℝ^{mtimes m}$ have nullity $m-l$?



The nullspace of $Q_lQ_l^T$ is {$x: Q_lQ_l^Tx=0$}



Since $Q_lQ_l^T=I$ why is the nullity not $0$?



Also if $A in ℝ^{mtimes n} , mge n$ why is $|Q_l^TA|_2 = |A|_2$ only if $l=m$?



I know that since $Q_lQ_l^T=I implies |Q_l|_2=1$ and so $|Q_l^TA|_2 le |A|_2$






linear-algebra




linear-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 21 '18 at 18:23







pablo_mathscobar

















asked Dec 21 '18 at 18:09









pablo_mathscobarpablo_mathscobar

1067




1067








  • 1




    $begingroup$
    The dimension of the nullspace is also called nullity
    $endgroup$
    – Shubham Johri
    Dec 21 '18 at 18:17










  • $begingroup$
    The nullity of $I$ is $0$
    $endgroup$
    – Shubham Johri
    Dec 21 '18 at 18:23










  • $begingroup$
    Was a typo my bad
    $endgroup$
    – pablo_mathscobar
    Dec 21 '18 at 18:24














  • 1




    $begingroup$
    The dimension of the nullspace is also called nullity
    $endgroup$
    – Shubham Johri
    Dec 21 '18 at 18:17










  • $begingroup$
    The nullity of $I$ is $0$
    $endgroup$
    – Shubham Johri
    Dec 21 '18 at 18:23










  • $begingroup$
    Was a typo my bad
    $endgroup$
    – pablo_mathscobar
    Dec 21 '18 at 18:24








1




1




$begingroup$
The dimension of the nullspace is also called nullity
$endgroup$
– Shubham Johri
Dec 21 '18 at 18:17




$begingroup$
The dimension of the nullspace is also called nullity
$endgroup$
– Shubham Johri
Dec 21 '18 at 18:17












$begingroup$
The nullity of $I$ is $0$
$endgroup$
– Shubham Johri
Dec 21 '18 at 18:23




$begingroup$
The nullity of $I$ is $0$
$endgroup$
– Shubham Johri
Dec 21 '18 at 18:23












$begingroup$
Was a typo my bad
$endgroup$
– pablo_mathscobar
Dec 21 '18 at 18:24




$begingroup$
Was a typo my bad
$endgroup$
– pablo_mathscobar
Dec 21 '18 at 18:24










1 Answer
1






active

oldest

votes


















1












$begingroup$

$Q_lQ_l^Tne I_m$. Observe however that $Q_l^TQ_l=I_l$, using that $Q$ was orthogonal. Thus the rank of $Q_l$ must be at least $l$, however since it is an $mtimes l$ matrix, it has rank at most $l$ at the same time.



Edit:
I mistakenly wrote $Q_l$ has nullity $m-l$. That was wrong. It has rank $l$ and therefore nullity $0$ by rank-nullity. It is $Q_l^T$ that has nullity $m-l$, also by rank-nullity (since matrices and their transposes have the same rank).



Then $Q_lQ_l^T$ has nullity $m-l$ since it is the product of an injective matrix and a matrix with nullity $m-l$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Why is the last line true?
    $endgroup$
    – pablo_mathscobar
    Dec 21 '18 at 18:54










  • $begingroup$
    @pablo_mathscobar Sorry, I was careless. I've edited to fix.
    $endgroup$
    – jgon
    Dec 21 '18 at 18:58










  • $begingroup$
    Why is $Q_l$ an injective matrix?
    $endgroup$
    – pablo_mathscobar
    Dec 21 '18 at 19:03










  • $begingroup$
    @pablo_mathscobar, because its nullity is 0
    $endgroup$
    – jgon
    Dec 21 '18 at 20:00










  • $begingroup$
    Oh thanks makes sense
    $endgroup$
    – pablo_mathscobar
    Dec 21 '18 at 20:11











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1












$begingroup$

$Q_lQ_l^Tne I_m$. Observe however that $Q_l^TQ_l=I_l$, using that $Q$ was orthogonal. Thus the rank of $Q_l$ must be at least $l$, however since it is an $mtimes l$ matrix, it has rank at most $l$ at the same time.



Edit:
I mistakenly wrote $Q_l$ has nullity $m-l$. That was wrong. It has rank $l$ and therefore nullity $0$ by rank-nullity. It is $Q_l^T$ that has nullity $m-l$, also by rank-nullity (since matrices and their transposes have the same rank).



Then $Q_lQ_l^T$ has nullity $m-l$ since it is the product of an injective matrix and a matrix with nullity $m-l$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Why is the last line true?
    $endgroup$
    – pablo_mathscobar
    Dec 21 '18 at 18:54










  • $begingroup$
    @pablo_mathscobar Sorry, I was careless. I've edited to fix.
    $endgroup$
    – jgon
    Dec 21 '18 at 18:58










  • $begingroup$
    Why is $Q_l$ an injective matrix?
    $endgroup$
    – pablo_mathscobar
    Dec 21 '18 at 19:03










  • $begingroup$
    @pablo_mathscobar, because its nullity is 0
    $endgroup$
    – jgon
    Dec 21 '18 at 20:00










  • $begingroup$
    Oh thanks makes sense
    $endgroup$
    – pablo_mathscobar
    Dec 21 '18 at 20:11
















1












$begingroup$

$Q_lQ_l^Tne I_m$. Observe however that $Q_l^TQ_l=I_l$, using that $Q$ was orthogonal. Thus the rank of $Q_l$ must be at least $l$, however since it is an $mtimes l$ matrix, it has rank at most $l$ at the same time.



Edit:
I mistakenly wrote $Q_l$ has nullity $m-l$. That was wrong. It has rank $l$ and therefore nullity $0$ by rank-nullity. It is $Q_l^T$ that has nullity $m-l$, also by rank-nullity (since matrices and their transposes have the same rank).



Then $Q_lQ_l^T$ has nullity $m-l$ since it is the product of an injective matrix and a matrix with nullity $m-l$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Why is the last line true?
    $endgroup$
    – pablo_mathscobar
    Dec 21 '18 at 18:54










  • $begingroup$
    @pablo_mathscobar Sorry, I was careless. I've edited to fix.
    $endgroup$
    – jgon
    Dec 21 '18 at 18:58










  • $begingroup$
    Why is $Q_l$ an injective matrix?
    $endgroup$
    – pablo_mathscobar
    Dec 21 '18 at 19:03










  • $begingroup$
    @pablo_mathscobar, because its nullity is 0
    $endgroup$
    – jgon
    Dec 21 '18 at 20:00










  • $begingroup$
    Oh thanks makes sense
    $endgroup$
    – pablo_mathscobar
    Dec 21 '18 at 20:11














1












1








1





$begingroup$

$Q_lQ_l^Tne I_m$. Observe however that $Q_l^TQ_l=I_l$, using that $Q$ was orthogonal. Thus the rank of $Q_l$ must be at least $l$, however since it is an $mtimes l$ matrix, it has rank at most $l$ at the same time.



Edit:
I mistakenly wrote $Q_l$ has nullity $m-l$. That was wrong. It has rank $l$ and therefore nullity $0$ by rank-nullity. It is $Q_l^T$ that has nullity $m-l$, also by rank-nullity (since matrices and their transposes have the same rank).



Then $Q_lQ_l^T$ has nullity $m-l$ since it is the product of an injective matrix and a matrix with nullity $m-l$.






share|cite|improve this answer











$endgroup$



$Q_lQ_l^Tne I_m$. Observe however that $Q_l^TQ_l=I_l$, using that $Q$ was orthogonal. Thus the rank of $Q_l$ must be at least $l$, however since it is an $mtimes l$ matrix, it has rank at most $l$ at the same time.



Edit:
I mistakenly wrote $Q_l$ has nullity $m-l$. That was wrong. It has rank $l$ and therefore nullity $0$ by rank-nullity. It is $Q_l^T$ that has nullity $m-l$, also by rank-nullity (since matrices and their transposes have the same rank).



Then $Q_lQ_l^T$ has nullity $m-l$ since it is the product of an injective matrix and a matrix with nullity $m-l$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 21 '18 at 18:58

























answered Dec 21 '18 at 18:35









jgonjgon

16k32143




16k32143












  • $begingroup$
    Why is the last line true?
    $endgroup$
    – pablo_mathscobar
    Dec 21 '18 at 18:54










  • $begingroup$
    @pablo_mathscobar Sorry, I was careless. I've edited to fix.
    $endgroup$
    – jgon
    Dec 21 '18 at 18:58










  • $begingroup$
    Why is $Q_l$ an injective matrix?
    $endgroup$
    – pablo_mathscobar
    Dec 21 '18 at 19:03










  • $begingroup$
    @pablo_mathscobar, because its nullity is 0
    $endgroup$
    – jgon
    Dec 21 '18 at 20:00










  • $begingroup$
    Oh thanks makes sense
    $endgroup$
    – pablo_mathscobar
    Dec 21 '18 at 20:11


















  • $begingroup$
    Why is the last line true?
    $endgroup$
    – pablo_mathscobar
    Dec 21 '18 at 18:54










  • $begingroup$
    @pablo_mathscobar Sorry, I was careless. I've edited to fix.
    $endgroup$
    – jgon
    Dec 21 '18 at 18:58










  • $begingroup$
    Why is $Q_l$ an injective matrix?
    $endgroup$
    – pablo_mathscobar
    Dec 21 '18 at 19:03










  • $begingroup$
    @pablo_mathscobar, because its nullity is 0
    $endgroup$
    – jgon
    Dec 21 '18 at 20:00










  • $begingroup$
    Oh thanks makes sense
    $endgroup$
    – pablo_mathscobar
    Dec 21 '18 at 20:11
















$begingroup$
Why is the last line true?
$endgroup$
– pablo_mathscobar
Dec 21 '18 at 18:54




$begingroup$
Why is the last line true?
$endgroup$
– pablo_mathscobar
Dec 21 '18 at 18:54












$begingroup$
@pablo_mathscobar Sorry, I was careless. I've edited to fix.
$endgroup$
– jgon
Dec 21 '18 at 18:58




$begingroup$
@pablo_mathscobar Sorry, I was careless. I've edited to fix.
$endgroup$
– jgon
Dec 21 '18 at 18:58












$begingroup$
Why is $Q_l$ an injective matrix?
$endgroup$
– pablo_mathscobar
Dec 21 '18 at 19:03




$begingroup$
Why is $Q_l$ an injective matrix?
$endgroup$
– pablo_mathscobar
Dec 21 '18 at 19:03












$begingroup$
@pablo_mathscobar, because its nullity is 0
$endgroup$
– jgon
Dec 21 '18 at 20:00




$begingroup$
@pablo_mathscobar, because its nullity is 0
$endgroup$
– jgon
Dec 21 '18 at 20:00












$begingroup$
Oh thanks makes sense
$endgroup$
– pablo_mathscobar
Dec 21 '18 at 20:11




$begingroup$
Oh thanks makes sense
$endgroup$
– pablo_mathscobar
Dec 21 '18 at 20:11


















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